This is in calculus using optimization.

The distance between two points in the xy plane is: D = [(x2-x1)^2 + (y2-y1)^2]^0.5 Let (x1,y1) be the point (3,0), then D is: [(x2-3)^2 + (y2)^2]^0.5, or to save clutter: D = [(x-3)^2 + y^2]^0.5 y = 8/x, so D = [(x-3)^2 + 64x^-2]^0.5 dD/dx = 0.5 [(x-3)^2 + 64x^-2]^-0.5 * [2(x-3) - 128x^-3] D is at the minimum when dD/dx = 0, so solve for: [2(x-3) - 128x^-3] = 0 x = 4, so y = 2. The hyperbola is closest to (3, 0) at (4, 2) where the distance is the square root of 5.

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